Find the largest constant $C$ so that
\[x^2 + y^2 + 1 \ge C(x + y)\]for all real numbers $x$ and $y.$
Solution: The given inequality expands as
\[x^2 + y^2 + 1 \ge Cx + Cy.\]Completing the square in $x$ and $y,$ we get
\[\left( x - \frac{C}{2} \right)^2 + \left( y - \frac{C}{2} \right)^2 + 1 - \frac{C^2}{2} \ge 0.\]This inequality holds for all $x$ and $y$ if and only if $1 - \frac{C^2}{2} \ge 0,$ or $C^2 \le 2.$  Thus, the largest possible value of $C$ is $\boxed{\sqrt{2}}.$